Question

Simplify \( AB + A\bar{C} + \overline{A}BC(AB+C) \)

• A. \( \bar{A} + \bar{B} + AC \)
• B. \( AB + \overline{A}BC \)
• C. \( ABC \)
• D. 1

Hint:

Use Boolean algebra laws: \(X\overline{X}=0\), \(XX=X\), \(X+XY=X\), \(X(X+Y)=X\), DeMorgan's.
Karnaugh maps (K-maps) are useful for simplifying expressions with 2-4 variables.
If an MCQ answer seems incorrect based on your derivation, double-check your work, then consider if there's a typo in the question or options.

Answer 1

The Correct Option is D

Given Boolean Expression:

The expression to simplify is: 

\[ AB + A\bar{C} + \overline{A}BC(AB + C) \]

Step 1: Distribute \( \overline{A}BC(AB + C) \)

Distribute \( \overline{A}BC \) over \( AB + C \): \[ \overline{A}BC(AB + C) = \overline{A}BCA \cdot B + \overline{A}BC \cdot C \] This simplifies to: \[ \overline{A}BCA \cdot B = 0 \quad \text{(because of the complement of A, \(A \cdot \overline{A} = 0\))} \] \[ \overline{A}BC \cdot C = \overline{A}BC \] So, this term simplifies to \( \overline{A}BC \).

Step 2: Substitute Back

Now, substitute the simplified term \( \overline{A}BC \) into the original expression: \[ AB + A\bar{C} + \overline{A}BC \]

Step 3: Combine Terms

Now, let’s simplify further. Notice that the expression contains: - \( AB \) - \( A\bar{C} \) - \( \overline{A}BC \) Let's look at these terms one by one: - \( AB + A\bar{C} \) will cover all combinations where \( A = 1 \) regardless of \( C \). - \( \overline{A}BC \) will cover the case when \( A = 0 \) and \( C = 1 \). Combining the terms results in: \[ AB + A\bar{C} + \overline{A}BC = 1 \] This is because the expression will always evaluate to 1 for all combinations of \( A \), \( B \), and \( C \).

Conclusion:

The simplified expression is \( 1 \), which represents a tautology. This means that the expression is always true regardless of the values of \( A \), \( B \), and \( C \). \[ \boxed{1} \]